I am building a DC-DC converter for a battery powered personal transportation unit. The battery is puttin out about 60V and 22A. I found a circuit design to take the voltage from 60V to about 5V ( which is what I need). But the circuit is only designed for 1.5A. Is there some way to fix this because I am not really sure what to do with such a high amperage. The circuit design and datasheet is attached as a pdf file.

 

What is your motors draw?

You would need a huge battery to sustain a 22 amp draw for any length of time.

 

5v for a motor big enough to power an electric vehicle?
I thought most were 12, 24 or 48v DC. If you haven't actually GOT a motor yet get a higher voltage one so you don't have to use thicker cables.

I'd suggest re-arranging the battery pack (no single battery is going to be 60v) to give a lower voltage and using a PWM controller to feed the motor.

 

The motor's draw is about 60V. The 5v is for a circuit to measure the speed and possibly an implementation of automatic lights. Im not working on the motor itself, I am simply adding things to the unit such as a speed gauge, watt meter, odometer, etc. Someone else on my team is dealing with the motor but I think we have 3 batteries hooked up to obtain to 60V.

 

Ok. So you are dealing with the 5v end. You want to use a voltage regulator to tap into the 60v line and get 5v to work with for the accessories, correct?

 

Yup, thats the issue right now.

 

Ok, a voltage divider to drop the 60v to 30v then a 7805 to get your 5v, unless Im missing something.

The 7805 has a max input of 35v, so the divider should do the trick. And as the batteries drain down, the voltage regulator will keep the 5v output constant.

Depending on the amount of devices you need to run, you could repeat this same circuit if you need more than the max current that the 7805 can provide.

 

@ retched

and this should work with the 22 Amps?

 

What is the current demand at 5v?

If it is low (say, 100mA), why don't you just run a 7805 regulator from the battery that's lowest in the series string? (closest to ground). That way you'll have 14v tops as input to the 7805.

Or, you could even use a switching regulator. Unless you need a lot of current at 5v, no point in going across the whole 60v pack.

 

The regulator wont get 22amps, It will only draw what in needs. You can have a 10000Ah battery and hook this up and it will work. It will work for about a thousand years.

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You could even do as SgtWookie said and run directly from the lowest battery in the string. If you have confusion about how this is done, you can still use the divider method. That way if you get it wrong, you will only be feeding 30v to the regulator, and if you get it right, you will be sending around 7v. This will be a little on the low side for the 7805. It is important to use your meter and judge your connections thusly.

From your request, Im guessing you will be using more than 100ma for the tach, speedo and the such, so either way. The thing about using the divider with the one battery method, if as the voltage drops in the battery, you may not be able to sustain the voltage to the 7805.
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I think what we have here is a failure to communicate.

You're asking about how to connect a small (in terms of power draw) system to a very large battery. If that's the case, there is no point in asking the question "What do I do with the 22A supply?"

It's like asking "Can I get a cup of water out of the Mississippi?" Yes, you can always take less than what's there. Your design needs to deal with the issue of stepping the 60V level down to 5V, and you say you've got that, and it can supply 1.5A at 5V, which sounds like plenty for an instrumentation circuit. So you're fine.

I hope that it's not a system with resistors and linear voltage regulators, though; that would be hideously inefficient running off 60V, and if you drew your full 1.5A off the battery, you'd be burning away over 70 Watts. You'd need to allow for some seriously large resistors or heat sinks to do that, and make sure you had good air circulation--all to waste battery power! Bad bad bad idea.

Details remaining are what happens if you short-circuit your 1.5A low-voltage supply during development? Is it protected against mechanical damage? And is there any possibility that your input could short out? With this much power available, you can have some quite spectacular accidents.